Using A as the origin (0,0), Let's drop a perpendicular FG to the X-axis and a perpendicular FH to the y-axis. Line AD is y= (1/2)x and Line EC is y = (9/8)x - 45 and they intersect at F(72,36). The pink area is made up of a 48 by 36 rectangle(FGBH), a 32 by 36 right triangle (EFG) and a 48 by 24 right triangle (DFH) so the area is 48*36 + (1/2)*32*36 + (1/2)*24*48 = 2880 square units.
Yes, you are using x-y coordinates and computing the coordinates of point F, after choosing point A as your origin and letting AB lie on the X axis in the right half plane. As you have calculated, the F coordinates are (72, 36). However, using PreMath's a and b, the coordinates of F are (120 - b, a), so PreMath's a = 36 and b = 48 and PreMath's formula pink area = 40a + 30b, presented at 11:37, may be used to compute the area.
I used a variation of this method. Put point A at (0,0) and then figure out where point F is in the coordinate system using the intersection of the two hypotenuses (F=72, 36). Then calculate the area of triangle ABD = 1/2*120*60 = 3600 and subtract the area of triangle AEF = 1/2*36*40 = 720, giving the total of the pink-shaded area = 3600-720= 2880 square units.
Let's use an orthonormal center E and first axis (EA) We have B(0; 0) E(80; 0) A(120; 0) D(0; 60) C(0; 90) We obtain easily the equation of (CE): 9.x +8.y -720 = 0 and of (AD): x +2.y -120 = 0, and their intersection F(48; 36) Then, area of triangle DBF = (1/2).DB.(abscissa of F) = (1/2).60.48 = 1440 And area of triangle BFE: (1/2).BE.(ordinate of F) = (1/2). 80. 36 = 1440 Finally, pink area = 1440 + 1440 = 2880.
Areas de triángulos: ABD=60*(80+40)/2=(60+30)*80/2=3600=CBE → Ambos triángulos tienen igual área→ Restándoles el solapo quedan dos triángulos sueltos de igual superficie→ AEF=CDF=a→ DBF=(60/30)a=2a=(80/40)a=EBF→ ABD=3600=5a→ a=720→ EBDF=4a=4*720=2880 ud². Bonito rompecabezas. Gracias y un saludo cordial.
Method using similar triangles, area ratios of 4 triangles in trapezoid divided by diagonals: 1. Join DE. Triangles BED and BAC are similar (ratio of 2 sides, inclusive angle) (equal side ratios of 2 pairs of corresponding sides are 80:(80+40) and 60:(60+30) = 2:3) Hence ED//AC and ED:AC = 2:3 2. Quadrilateral ACDE is trapezoid as ED//AC with top side ED:bottom side AC = 2:3 Area ratios of the 4 triangles in the trapezoid DEF:AFE:CDF:ACF = 4:6:6:9 Actual area of trapezoid = area of triangle BAC - area of triangle BED = 5400 - 2400 = 3000 Hence actual area of triangle DEF = [4/(9+6+6+4)](3000) = 480 3. Area of pink region = area of triangle BED + area of triangle DEF = 2400 + 480 = 2880
B=(0,0) 1) Hallar las coordenadas de F 2) Trazar el segmento FB, que divide al cuadrilátero EFDB en 2 triangulos T1= EFDB. T2= FBD 3) Hallar las AREAS de los triangulos 4) área cuadrilátero = área T1+área T2 =2*1440 MI RESPUESTA DESDE Bogotá D.C. COLOMBIA
I found a way which is simpler, but it uses analytic geometry. Here are the steps: 1) Set point A as the origin (x=0, y=0) 2) Make equations for lines AD and EC 3) Solve the two equations to find x and y. 4) The y value is the same as your "a". Your "b" value is 120-x 5) Add up the areas of the EFB and BFD triangles For line AD, y = x/2. For line EC, y = 9x/8 - 45 (using point/slope method). Setting the two equations equal, x = 72. Since y = x/2, y = 36. Your "a" value is 36 and "b" is 120-72 = 48. The two triangle areas added are 36*80/2 + 48*60/2 = 2880.
Let the Coordinates of Point B = (0 ; 0) Let the Coordinates of Point F = (x ; y) Equation of Straight Line AD : x / 120 + y / 60 = 1 Equation of Straight Line EC : x / 80 + y / 90 = 1 Solving the following System of 2 Linear Equations with 2 Unkowns : x / 120 + y / 60 = 1 x / 80 + y / 90 = 1 Solutions : x = 48 and y = 36 2 * Area = (60 * x) + (80 * y) ; 2 * Area = (60 * 48) + (80 * 36) ; 2 * A = 2.880 + 2.880 ; A = 2.880 Pink Area = 2.880 Square Units
Assign point B (0,0) Eqn for Line AD: Y=(90/80)X+90 or Y=9/8X+90 Eqn for Line EC: Y=(60/120)X+60 or Y=1/2X+60 Find intersection (pt F): 1/2X+60=9/8X+90 X=(-48); Y=36 Area= Rectangle (48x36)=1728 Triangle 1/2(80-48)(36)=576 Triangle 1/2(48)(60-36)=576 Total Area: 2880
1/ Label FT=a and ET=x Area = Area [EBD]+Area[EFD]= Area[EBD]+Area[AED]-Area[AEF] =2400+1200-Area [AEF] =3600-Area[AEF] 2/The 2 triangles AFT and ABD are similar so: FT/ET=DB/AB=60/120=1/2 --> a/(40+x) =1/2-> x=2a-40 (1) 2/ The 2 triangles EFT snd ECD are similar so x/a= EB/BC=80/90= 8/9 (2) From (1) and (2) we have: (2a-40)/a = 8/9-> a=36 --> Area= 3600 - 1/2 . 36. 40 = 3600- 720= 2880 sq units
Draw BF. By observation, it is apparent that the pink area is equal to the sum of the areas of the triangles ∆BFE and ∆DFB. Drop perpendiculars from F to M on AB and N on BC. Let FM = y and FN = x. Triangle ∆ABD: A₁ = ∆ABF + ∆DFB 120(60)/2 = 120y/2 + 60x/2 3600 = 60y + 30x ---- [1] Triangle ∆EBC: A₂ = ∆BFE + ∆FBC 90(80)/2 = 80y/2 + 90x/2 3600 = 40y + 45x ---- [2] Multiply [2] by 3/2 and subtract from [1]. 3600 - 3600(3/2) = 60y + 30x - (40y+45x)(3/2) 3600 - 5400 = 60y + 30x - 60y - 135x/2 -1800 = -75x/2 75x = 3600 x = 3600/75 = 48 3600 = 60y + 30(48)
It’s so simple. Areas of AEF and BEF 1 : 2, also CDF:BDF 1 : 2. Area of ABD = area of BCE = 3600. So ABD consists of x + 2x + 2x, x = 3600/5 = 720, pink = 4 * 720 = 2880.
Here's how I solved: Join B to F to form 4 triangles. △AEF and △BEF have same height (altitude from F to AB) △AEF has base = 40 and △BEF has base = 80, so it has twice area of △AEF Area(△AEF) = x, Area(△BEF) = 2x △BDF and △CDF have same height (altitude from F to BC) △CDF has base = 30 and △BDF has base = 60, so it has twice area of △CDF Area(△CDF) = y, Area(△BDF) = 2y Area(△ABD) = Area(△AEF) + Area(△BEF) + Area(△BDF) 1/2(120)(60) = x + 2x + 2y *3x + 2y = 3600 (1)* Area(△BCE) = Area(△BEF) + Area(△BDF) + Area(△CDF) 1/2 (80)(90) = 2x + 2y + y *2x + 3y = 3600 (2)* Adding (1) and (2) we get 5x + 5y = 7200 *x + y = 1440* Area of pink quadrilateral = Area(△BEF) + Area(△BDF) = 2x + 2y = 2(x+y) = 2(1440) *= 2880*
once you know that Area ABD = Area EBC = 3600 you can easily show that area AEF = area FDC, then 1/2*40*a = 1/2*30*b b = 4/3 a triangles AFT and ABD are similar then a : 60 = (120 - 4/3a) : 120 120a = 7200 - 240/3a a = 36 Pink area = 3600 - 1/2*40*36 = 3600 - 720 = 2880
I used functions. y1 = x/2 y2 = 9/8(x-40) Intersection at x = 72 Area = integral(9/8(x-40)dx)+integral(x/2dx) with the boundaries of integration being x =40 to 72 and x = 72 to 120, respectively.
There’s a much easier solution. Join pb as shown, but notice that triangles AFE and EFB have the same height but their bases are in a ratio of 2 to 1. Therefore their areas are also in the same ratio. Likewise for the other pair of triangles at the top right of the diagram. Using these facts you can quickly determine the pink area.
Se traza altura con dirección horizontal Por relación de lados del triangulo vertical, la altura mide 9a y su base horizontal 8a. Aplicamos la relación de lados del triangulo horizontal de 1 a 2, por tanto 18a = 40 + 8a a = 4 La altura de la intersección a la base mide 36 Trazamos una paralela a la base desde la intersección qué mide: 80 - 32 = 48 Suma de Áreas los 2 triángulos + rectángulo Área del rectángulo 48x36 = 1728 Área del triangulo de la intersección a la base = 32x36/2 = 576 Área del triangulo de la intersección a la vertical = 48x24/2 = 576 1728 + 576 576 Area = 2880
Hello, can you show us your book of exercice please. It is long time I follow your channel but I need to know where do you find these amazing exercice.
Nice. I solved for the (x,y) coordinates of point F by intersecting the two lines: AD & EC after calculating their equations from the slope of each given the provided dimensions and relevant points, and then proceeded to compute the area.
Del quadrilatero conosco i4 angoli e 2 lati .. perciò conosco l'area..i due lati mancanti sono 20√29/√5 e 120/√5...perciò l'area è la somma di 2400+480=2880
That’s very difficult ..
But the explanation is good .
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Using A as the origin (0,0), Let's drop a perpendicular FG to the X-axis and a perpendicular FH to the y-axis. Line AD is y= (1/2)x and Line EC is y = (9/8)x - 45 and they intersect at F(72,36). The pink area is made up of a 48 by 36 rectangle(FGBH), a 32 by 36 right triangle (EFG) and a 48 by 24 right triangle (DFH) so the area is 48*36 + (1/2)*32*36 + (1/2)*24*48 = 2880 square units.
Yes, you are using x-y coordinates and computing the coordinates of point F, after choosing point A as your origin and letting AB lie on the X axis in the right half plane. As you have calculated, the F coordinates are (72, 36). However, using PreMath's a and b, the coordinates of F are (120 - b, a), so PreMath's a = 36 and b = 48 and PreMath's formula pink area = 40a + 30b, presented at 11:37, may be used to compute the area.
Good one. A quick way to find the coordinates of F (a and b). I set the origin at B; the y intercepts for AD and EC are given to you.
I used a variation of this method. Put point A at (0,0) and then figure out where point F is in the coordinate system using the intersection of the two hypotenuses (F=72, 36). Then calculate the area of triangle ABD = 1/2*120*60 = 3600 and subtract the area of triangle AEF = 1/2*36*40 = 720, giving the total of the pink-shaded area = 3600-720= 2880 square units.
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Let's use an orthonormal center E and first axis (EA)
We have B(0; 0) E(80; 0) A(120; 0) D(0; 60) C(0; 90)
We obtain easily the equation of (CE): 9.x +8.y -720 = 0
and of (AD): x +2.y -120 = 0,
and their intersection F(48; 36)
Then, area of triangle DBF = (1/2).DB.(abscissa of F)
= (1/2).60.48 = 1440
And area of triangle BFE: (1/2).BE.(ordinate of F)
= (1/2). 80. 36 = 1440
Finally, pink area = 1440 + 1440 = 2880.
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ABD=120*60/2=3600 EBC=90*80/4=3600
40 : 80 = 1: 2 30 : 60 = 1: 2
AEF=x FEB=2x CDF=y BFD=2y
x+2x+2y=3600 y+2y+2x=3600 3x+2y=2x+3y x=y
5x=3600 x=720 y=720
Pink Area = 2x+2y = 2(x+y) = 2*1440 = 2880
I did something similar to this.
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Areas de triángulos: ABD=60*(80+40)/2=(60+30)*80/2=3600=CBE → Ambos triángulos tienen igual área→ Restándoles el solapo quedan dos triángulos sueltos de igual superficie→ AEF=CDF=a→ DBF=(60/30)a=2a=(80/40)a=EBF→ ABD=3600=5a→ a=720→ EBDF=4a=4*720=2880 ud².
Bonito rompecabezas. Gracias y un saludo cordial.
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Method using similar triangles, area ratios of 4 triangles in trapezoid divided by diagonals:
1. Join DE. Triangles BED and BAC are similar (ratio of 2 sides, inclusive angle)
(equal side ratios of 2 pairs of corresponding sides are 80:(80+40) and 60:(60+30) = 2:3)
Hence ED//AC and ED:AC = 2:3
2. Quadrilateral ACDE is trapezoid as ED//AC with top side ED:bottom side AC = 2:3
Area ratios of the 4 triangles in the trapezoid DEF:AFE:CDF:ACF = 4:6:6:9
Actual area of trapezoid = area of triangle BAC - area of triangle BED = 5400 - 2400 = 3000
Hence actual area of triangle DEF = [4/(9+6+6+4)](3000) = 480
3. Area of pink region = area of triangle BED + area of triangle DEF = 2400 + 480 = 2880
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B=(0,0)
1) Hallar las coordenadas de F
2) Trazar el segmento FB, que divide al cuadrilátero EFDB en 2 triangulos T1= EFDB. T2= FBD
3) Hallar las AREAS de los triangulos
4) área cuadrilátero =
área T1+área T2 =2*1440 MI RESPUESTA
DESDE Bogotá D.C. COLOMBIA
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Dropping perpendicular from point F onto EB to point G.
Let length FG = h.
Let length EG = x.
Triangles AFG & ADB are similar.
So h / (40 + x) = 60 /120.
h = (40 + x) / 2.
Triangles EFG & ECB are similar.
So x / h = 80 / 90.
h = 9x / 8.
Equating 2 values for h.
(40 + x) / 2 = 9x / 8.
18x = 320 + 8x.
10x =320
x = 32.
h= 9x / 8 (as previously calculated) = 9 * 32 / 8 = 36.
Area triangle AEF = 1/2 * 40 * 36 = 720.
Area triangle ABD = 1/2 * 120 * 60 = 3600.
Pink area = 3600 - 720 = 2880.
line 1: y = 60/120 x + 60 = 1/2 x + 60
line 2: y = 90/80 x + 90 = 9/8 x + 90
y = y
1/2 x + 60 = 9/8 x + 90
(9/8 - 1/2) x = 60 - 90
(9/8 - 4/8) x = - 30
5/8 x = - 30
x = -30 * 8/5 = - 6 * 8 = - 48
y = 1/2 * - 48 + 60 = - 24 + 60 = 36
A(red) = 1/2 (80 * y + 60 * |x| = 1/2 (80 * 36 + 60 * 48) = 40 * 36 + 30 * 48 = 2880 square units
I found a way which is simpler, but it uses analytic geometry. Here are the steps:
1) Set point A as the origin (x=0, y=0)
2) Make equations for lines AD and EC
3) Solve the two equations to find x and y.
4) The y value is the same as your "a". Your "b" value is 120-x
5) Add up the areas of the EFB and BFD triangles
For line AD, y = x/2. For line EC, y = 9x/8 - 45 (using point/slope method). Setting the two equations equal, x = 72. Since y = x/2, y = 36. Your "a" value is 36 and "b" is 120-72 = 48. The two triangle areas added are 36*80/2 + 48*60/2 = 2880.
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Let the Coordinates of Point B = (0 ; 0)
Let the Coordinates of Point F = (x ; y)
Equation of Straight Line AD : x / 120 + y / 60 = 1
Equation of Straight Line EC : x / 80 + y / 90 = 1
Solving the following System of 2 Linear Equations with 2 Unkowns :
x / 120 + y / 60 = 1
x / 80 + y / 90 = 1
Solutions : x = 48 and y = 36
2 * Area = (60 * x) + (80 * y) ; 2 * Area = (60 * 48) + (80 * 36) ; 2 * A = 2.880 + 2.880 ; A = 2.880
Pink Area = 2.880 Square Units
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Very smart, many thanks, Sir!
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Assign point B (0,0)
Eqn for Line AD: Y=(90/80)X+90 or
Y=9/8X+90
Eqn for Line EC: Y=(60/120)X+60 or
Y=1/2X+60
Find intersection (pt F):
1/2X+60=9/8X+90
X=(-48); Y=36
Area=
Rectangle (48x36)=1728
Triangle 1/2(80-48)(36)=576
Triangle 1/2(48)(60-36)=576
Total Area: 2880
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1/ Label FT=a and ET=x
Area = Area [EBD]+Area[EFD]= Area[EBD]+Area[AED]-Area[AEF]
=2400+1200-Area [AEF]
=3600-Area[AEF]
2/The 2 triangles AFT and ABD are similar so:
FT/ET=DB/AB=60/120=1/2
--> a/(40+x) =1/2-> x=2a-40 (1)
2/ The 2 triangles EFT snd ECD are similar so
x/a= EB/BC=80/90= 8/9 (2)
From (1) and (2) we have:
(2a-40)/a = 8/9-> a=36
--> Area= 3600 - 1/2 . 36. 40 = 3600- 720= 2880 sq units
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AE : EB = CD : DB = 1 : 2, ⇛, AC//DE, ACDE - тrapezoid.
Additional construction: FH ⟂ AC, FK ⟂ DE. FK : FH = DE : AC = 100 : 150 = 2 : 3.
S(ABC) = 5400, S(BDE) = 2400, S(ACDE) = 3000. HK = 3000/[0,5(AC + DE) = 3000/[0,5(150 + 100) = 24.
FK = (2/5) × HK = (2/5) × 24 = 9,6. S(EFD) = 1/2(DE × FK) = 1/2(100 × 9,6) = 480.
S(pink) = S(BDE) + S(EFD) = 2400 + 480 = 2880.
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Let's find the area:
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..
...
....
.....
Let's assume that B is the center of the coordinate system and that BA and BC are located on the x-axis and y-axis, respectively. Then we obtain the following coordinates from the sketch:
A: ( −120 ; 0 )
B: ( 0 ; 0 )
C: ( 0 ; 90 )
D: ( 0 ; 60 )
E: ( −80 ; 0 )
F: ( xF ; yF )
The lines AD and EC are represented by the follwing functions:
AD: y = (yD − yA)*(x − xA)/(xD − xA) = (60 − 0)*(x + 120)/(0 + 120) = (x + 120)/2
EC: y = (yC − yE)*(x − xE)/(xC − xE) = (90 − 0)*(x + 80)/(0 + 80) = 9*(x + 80)/8
These two lines intercept at F. Therefore we obtain:
(xF + 120)/2 = 9*(xF + 80)/8
4*(xF + 120) = 9*(xF + 80)
4*xF + 480 = 9*xF + 720
−5*xF = 240
⇒ xF = −48
yF = (xF + 120)/2 = (−48 + 120)/2 = 72/2 = 36
yF = 9*(xF + 80)/8 = 9*(−48 + 80)/8 = 9*32/8 = 9*4 = 36 ✓
A(pink)
= A(ABD) − A(AEF)
= (1/2)*AB*h(AB) − (1/2)*AE*h(AE)
= (1/2)*AB*BD − (1/2)*AE*yF
= (1/2)*120*60 − (1/2)*40*36
= 3600 − 720
= 2880
A(pink)
= A(BCE) − A(CDF)
= (1/2)*BC*h(BC) − (1/2)*CD*h(CD)
= (1/2)*BC*BE − (1/2)*CD*(−xF)
= (1/2)*90*80 − (1/2)*30*48
= 3600 − 720
= 2880 ✓
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Draw BF. By observation, it is apparent that the pink area is equal to the sum of the areas of the triangles ∆BFE and ∆DFB. Drop perpendiculars from F to M on AB and N on BC. Let FM = y and FN = x.
Triangle ∆ABD:
A₁ = ∆ABF + ∆DFB
120(60)/2 = 120y/2 + 60x/2
3600 = 60y + 30x ---- [1]
Triangle ∆EBC:
A₂ = ∆BFE + ∆FBC
90(80)/2 = 80y/2 + 90x/2
3600 = 40y + 45x ---- [2]
Multiply [2] by 3/2 and subtract from [1].
3600 - 3600(3/2) = 60y + 30x - (40y+45x)(3/2)
3600 - 5400 = 60y + 30x - 60y - 135x/2
-1800 = -75x/2
75x = 3600
x = 3600/75 = 48
3600 = 60y + 30(48)
It’s so simple. Areas of AEF and BEF 1 : 2, also CDF:BDF 1 : 2. Area of ABD = area of BCE = 3600. So ABD consists of x + 2x + 2x, x = 3600/5 = 720, pink = 4 * 720 = 2880.
Here's how I solved:
Join B to F to form 4 triangles.
△AEF and △BEF have same height (altitude from F to AB)
△AEF has base = 40 and △BEF has base = 80, so it has twice area of △AEF
Area(△AEF) = x, Area(△BEF) = 2x
△BDF and △CDF have same height (altitude from F to BC)
△CDF has base = 30 and △BDF has base = 60, so it has twice area of △CDF
Area(△CDF) = y, Area(△BDF) = 2y
Area(△ABD) = Area(△AEF) + Area(△BEF) + Area(△BDF)
1/2(120)(60) = x + 2x + 2y
*3x + 2y = 3600 (1)*
Area(△BCE) = Area(△BEF) + Area(△BDF) + Area(△CDF)
1/2 (80)(90) = 2x + 2y + y
*2x + 3y = 3600 (2)*
Adding (1) and (2) we get
5x + 5y = 7200
*x + y = 1440*
Area of pink quadrilateral
= Area(△BEF) + Area(△BDF)
= 2x + 2y = 2(x+y) = 2(1440)
*= 2880*
once you know that Area ABD = Area EBC = 3600 you can easily show that area AEF = area FDC, then
1/2*40*a = 1/2*30*b
b = 4/3 a
triangles AFT and ABD are similar then
a : 60 = (120 - 4/3a) : 120
120a = 7200 - 240/3a
a = 36
Pink area = 3600 - 1/2*40*36 = 3600 - 720 = 2880
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I used functions.
y1 = x/2
y2 = 9/8(x-40)
Intersection at x = 72
Area = integral(9/8(x-40)dx)+integral(x/2dx) with the boundaries of integration being x =40 to 72 and x = 72 to 120, respectively.
[BEFD] = [ADB] - [AFE]. Now [AFE] = AE*height/2. Now X/2 = (X-40) * 9/8 -> 40*9/8 = (9/8-1/2) X -> 45=5X/8 -> X=72 -> height = X/2 = 36. Now [AFE] = 40*36/2 = 720 and [ADB] = 120*60/2 = 3600 --> [BEFD] = 3600-720 = 2880 sq. units
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There’s a much easier solution. Join pb as shown, but notice that triangles AFE and EFB have the same height but their bases are in a ratio of 2 to 1. Therefore their areas are also in the same ratio. Likewise for the other pair of triangles at the top right of the diagram. Using these facts you can quickly determine the pink area.
Se traza altura con dirección horizontal
Por relación de lados del triangulo vertical, la altura mide 9a y su base horizontal 8a.
Aplicamos la relación de lados del triangulo horizontal de 1 a 2, por tanto
18a = 40 + 8a
a = 4
La altura de la intersección a la base mide 36
Trazamos una paralela a la base desde la intersección qué mide:
80 - 32 = 48
Suma de Áreas los 2 triángulos + rectángulo
Área del rectángulo 48x36 = 1728
Área del triangulo de la intersección a la base = 32x36/2 = 576
Área del triangulo de la intersección a la vertical = 48x24/2 = 576
1728 +
576
576
Area = 2880
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Yes, I got it!
Since triangles AFE and CFD have the same area ,it is easy to establish that b= (4/3)a ,so you can do it using just one variable.
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Hello, can you show us your book of exercice please. It is long time I follow your channel but I need to know where do you find these amazing exercice.
Nice. I solved for the (x,y) coordinates of point F by intersecting the two lines: AD & EC after calculating their equations from the slope of each given the provided dimensions and relevant points, and then proceeded to compute the area.
Thanks for the feedback ❤️
Why did you multiply by -20 for the formula 2?
Now I've got pink eye! 🙂
😀
pink eye😀
Del quadrilatero conosco i4 angoli e 2 lati .. perciò conosco l'area..i due lati mancanti sono 20√29/√5 e 120/√5...perciò l'area è la somma di 2400+480=2880
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Always geometry!
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There's something strange about this problem , if I use trigonometry I get 2852
A=1/2×90×120=5400, a+b=1/3×5400=1800, c+d=3600, b+d=3600, a+c=1800, seems not sufficient to solve for d?😢😢😢
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asnwer=125 is it what why math matter
asnwer=2880cm isit what
Thanks for the feedback ❤️